Figure shows two cylinders of radii $r_1$ and $r_2$ having moments of inertia $I_1$ and $I_2$ about their respective axes. Initially, the cylinders rotate about their axes with angular speed ${\omega }_1$ and ${\omega }_2$, as shown in the figure. The cylinders are moved closer to touch each other keeping the axes parallel. The cylinders first slip over each other at the contact but the slipping finally ceases due to the friction between them. Find the angular speeds of the cylinders after the slipping ceases.

In the process, frictional force at contact point consitutes torque, which causes change in angular momentum of the cylinders. If ${\omega }_1\text{'}$ and ${\omega }_2\text{'}$ are the final angular velocities of the cylinder then for no
slipping

                         ${\omega }_1\text{'} r_1 ={\omega }_2\text{'} r_2$             ......(i)

For first cylinder           $– f r_1 t = I_1 \left({\omega }_1\text{'}-{\omega }_1\right)$          .....(ii)

For second cylinder      $f r_2 t = I_2 \left({\omega }_2\text{'}-{\omega }_2\right)$          .....(iii)

After solving above equations, we get 

             ${\omega }_1\text{'} = \left[\frac{I_1{\omega }_1{r}_2 + I_2{\omega }_2{r}_1}{I_2{r_1}^2 + I_1{r_2}^2 }\right]r_2$

and     ${\omega }_2\text{'} = \left[\frac{I_1{\omega }_1{r}_2 + I_2{\omega }_2{r}_1}{I_2{r_1}^2 + I_1{r_2}^2 }\right]r_1$